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Example 30 - Find coordinates of point where line through A (3, 4, 1)

Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5


Transcript

Example 30 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points with position vectors π‘Ž βƒ— & 𝑏 βƒ— is 𝒓 βƒ— = 𝒂 βƒ— + πœ† (𝒃 βƒ— βˆ’ 𝒂 βƒ—) Given the line passes through the points (𝑏 βƒ— βˆ’ π‘Ž βƒ—) = (5𝑖 Μ‚ + 1𝑗 Μ‚ + 6π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚) = 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚ A (3, 4, 1) π‘Ž βƒ— = 3𝑖 Μ‚ + 4𝑗 Μ‚ + π‘˜ Μ‚ B (5, 1, 6) 𝑏 βƒ— = 5𝑖 Μ‚ + 1𝑗 Μ‚ + 6π‘˜ Μ‚ ∴ π‘Ÿ βƒ— = (3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚) + πœ† (2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, π‘Ÿ βƒ— = x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ = 3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚ + 2πœ†π‘– Μ‚ βˆ’ 3πœ†π‘— Μ‚ + 5πœ†π‘˜ Μ‚ x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ = (3 + 2πœ†)𝑖 Μ‚ + (4 βˆ’ 3πœ†)𝑗 Μ‚ + (1 + 5πœ†)π‘˜ Μ‚ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5πœ† ∴ πœ† = (βˆ’πŸ)/πŸ“ So, x = 3 + πœ† = 3 + 2 Γ— (βˆ’1)/5 = 3 βˆ’ 2/5 = 13/5 & y = 4 βˆ’ 3πœ† = 4 βˆ’ 3 Γ— (βˆ’1)/5 = 4 + 3/5 = 23/5 Therefore, the required coordinates are (πŸπŸ‘/πŸ“,πŸπŸ‘/πŸ“,𝟎) Example 30 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points A(π‘₯_1, 𝑦_1, 𝑧_1) and B(π‘₯_2, 𝑦_2, 𝑧_2) is (𝒙 βˆ’ 𝒙_𝟏)/(𝒙_𝟐 βˆ’ 𝒙_𝟏 ) = (π’š βˆ’ π’š_𝟏)/(π’š_𝟐 βˆ’ π’š_𝟏 ) = (𝒛 βˆ’ 𝒛_𝟏)/(𝒛_𝟐 βˆ’ 𝒛_𝟏 ) Given the line passes through the points So, the equation of line is (π‘₯ βˆ’ 3)/(5 βˆ’ 3) = (𝑦 βˆ’ 4)/(1 βˆ’ 4) = (𝑧 βˆ’ 1)/(6 βˆ’ 1) A (3, 4, 1) ∴ π‘₯_1= 3, 𝑦_1= 4, 𝑧_1= 1 B(5, 1, 6) ∴ π‘₯_2 = 5, 𝑦_2= 1, 𝑧_2= 6 (π‘₯ βˆ’ 3)/2 = (𝑦 βˆ’ 4)/(βˆ’3) = (𝑧 βˆ’ 1)/5 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = βˆ’1 ∴ k = (βˆ’πŸ)/πŸ“ So, x = 2k + 3 = 2 Γ— (βˆ’1)/5 + 3 = 3 βˆ’ 2/5 = 13/5 y = βˆ’3 Γ— (βˆ’1)/5 + 4 = 4 + 3/5 = 23/5 Therefore, the coordinates of the required point are (πŸπŸ‘/πŸ“, πŸπŸ‘/πŸ“, 𝟎).

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.